Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations KootK on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

GD & T Help needed please!

Status
Not open for further replies.

UGMENTALCASE

Aerospace
Oct 10, 2011
123
Good evening all,

I've been researching GD & T and purchased some good reference books which have helped me understand a lot, however I think I'm confusing myself a bit. In examples in the books I've been reading you have a pin and a hole. Hole is tolerance 12.0 + 0.1 - 0.1, so we calculate MMC and similarly with the shaft/pin. Calculate the VC and that's it, the parts fit together.
I'm struggling to find a real life scenario, so to speak. I've attached an image the blue outline is a profile, the grey inner track is a checking pin shape. Don't ask why it's just this shape (I have a similar situation I'm looking at) The line tolerance is stated, and there are some boxed dimensions. As far as I can see the only thing defining the line, is the profile tolerance, there are no ± dimensions anywhere etc. So if I were to want a fit of something like a g6 in this profile, could someone please help me out with how you would go about calculating it please?
I've gone over it as per basic examples (briefly outlined above) but I some how convince myself I'm doing it wrong, though I don't think I am! I've put an example of what I'm looking at on the end of the attachment. I hope this comes across ok, the example on the attachment is just a random tolerance to show how the hole and pin fit together, it's not using actual fits. I think I'm missing something somewhere but like I say I keep going round in circles!

Thanks in Advance
 
 http://files.engineering.com/getfile.aspx?folder=1c59b90e-92a3-4a1c-a2b1-db7818f2c3a5&file=04011601.PDF
Replies continue below

Recommended for you

UGMENTALCASE,

On your sketch, the profile tolerance completely controls your outline. The profile controls the size of your hole, and its position with respect to your datums. There is no need for a [±][ ]tolerance.

Read up on profile tolerances.

Your profile tolerance of 1mm imposes a width tolerance of [±]1mm. This is way too sloppy for a tolerance that mates with[ ]g6. The Machinery's Handbook shows H7. I would argue that you could apply a width of 16.018/16 to your slot. The profile would control the location of your slot. There was a discussion on mixing profiles with [±][ ]tolerances in thread1103-400077 in which I was disagreed with. You should read through it carefully.

--
JHG
 
Wow that's quite a thread! Well the slot/hole is set in stone. Let's say that's the component so I can't change that, the bit I just need my pin to fit in. And I suppose in a sense it leads to another questions, how do you tolerance the fit on something as sloppy as that? If the fit between them was better, I'm thinking of something like a 12 DIA hole H7 g6, the fit comes straight out of a book, and it will work. But if the tolerance is dictated by component drawings (in this instance) what do you aim for?
 
UGMENTALCASE,

Given that your outline is controlled by the profile tolerance, your maximum clearance cannot be less than a millimeter. For one piece only, you can measure the slot, and fabricate a pin to fit. For a batch of interchangeable parts, you can make a 10mm pin and systematically pick up one side of the slot. This will not be accurate, but it will be repeatable.

--
JHG
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor